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I have been working on this problem for a while now and think I need assistance. I am trying to solve with respect to $x_{(t)}$ over the interval $t = [0, \infty]$:

$$(m_{(t)} x')' + kx = 0$$ $$m_{(t)} = \frac{m_o}{1 + \alpha t}$$

Subject to he boundary conditions:

$$x(o) = x_o $$ $$\dot x(0) = 0$$

This is clearly of the Sturm - Liouville form which is unweighted as the weight function $P_{(t)} = 1$

$(r(x)y')' + \left( q(x) + \lambda P(x) \right)y = 0$

$k, \alpha, m_o$ are all constants, and I know that the solution Involves Bessel functions so I have been trying to get the Differential Equation into the form of a Bessel equation:

$$(x^ry')' + (ax^s + bx^{r-2})y = 0$$

Which has the solution if $(1 - r)^2 > = 4b$ and either $s >(r-2)$ or $b = 0$

$$y = x^\alpha \left[ C_1 J_\nu( \lambda x^ \gamma) + C_2 Y_\nu(\lambda x^ \gamma) \right]$$

where $\alpha = \frac{1 - r}{2}$ $\gamma = \frac{2 - r +s}{2}$ $\lambda = 2 \frac{\sqrt{a}}{2-r+s}$ $\nu = \frac{ \sqrt{(1 - r)^2 - 4b}}{2 - r +s}$

I am not sure if this is the right road to go down because I have tried just about everything I can think off to solve this thing. A poke in the right direction would not be soon forgotten. Thank you for your time.

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Not sure how one gets Bessels from your equation, as it is a 1st order equation. Are you sure the 2nd term is $k x'$, or is it $k x$? – Ron Gordon Jan 12 '13 at 3:04
your right i made a mistake – Cactus BAMF Jan 12 '13 at 3:14
up vote 1 down vote accepted

You can simplify things a bit by setting $u=1+\alpha t$ and rearranging things a bit to get

$$\left [ u^{-1} y' \right ]' + \frac{k}{m_0 \alpha^2} y = 0 $$

where $y(u) = x \left ( \frac{u-1}{\alpha} \right ) = x(t) $. You can map this equation (which is really not much more than what you already have) onto the parameter values $r=-1$, $s=0$, $a =\frac{k}{m_0 \alpha^2}$, and $b=0$. You should be able to take it form here.

In passing, I note that the index $\nu = \frac{2}{3}$ with an argument of that Bessel being raised to the power $\frac{3}{2}$ is indicative of an Airy function, which you may wish to investigate a little.

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Thank you very much, I had started to make the substitution you have there, but I did not see any use in it at the time. – Cactus BAMF Jan 12 '13 at 4:31

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